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Math Help - Need help with a formula!

  1. #1
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    Need help with a formula!

    I'm not sure if this is the correct forum for this problem, but I am puzzled by this. I'm not the best with maths so I don't know how to solve this.
    There are three numbers (below) that make three results (also below). Thing is I don't know if the 3rd number is required or not. There is a formula for each of the results. I know there is a formula for the result (each result 1 has the same formula, each result 2 has the same formula and each result 3 has the same formula), but I don't know if there is any other hidden variables to it or not. All I have figured out is that Number 2 is 25 times the value of number 1. Can anyone solve this with the information here?

    Number 1: 76
    Number 2: 1900
    Number 3: 152
    Result 1: 95.4
    Result 2: 31.4
    Result 3: 24.8

    Number 1: 106
    Number 2: 2650
    Number 3: 212
    Result 1: 192
    Result 2: 63.3
    Result 3: 49.8

    Number 1: 112
    Number 2: 2800
    Number 3: 224
    Result 1: 221.7
    Result 2: 73.2
    Result 3: 57.6

    Number 1: 132
    Number 2: 3300
    Number 3: 264
    Result 1: 361.5
    Result 2: 119.2
    Result 3: 93.8
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  2. #2
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    Re: Need help with a formula!

    For the function that takes Number 1 -> Result 1, the following quartic equation works:

    x^4 - \dfrac{1116459377}{2620800}x^3 + \dfrac{12588804713}{187200}x^2 - \dfrac{1525417666789}{327600}x + \dfrac{232244321233}{1950}.

    I don't have time to find quartics for the other two sets of numbers.
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  3. #3
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    Re: Need help with a formula!

    I think you misunderstood my question (or I didn't clarify myself). Number 1 2 and (possibly) 3 are used to make result 1 and another formula to make result 2 and another one to make result 3
    Last edited by Howlin1; October 18th 2013 at 11:39 AM.
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  4. #4
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    Re: Need help with a formula!

    Call number 1 x, number 2 y, number 3 z
    As you have pointed out already (\text{number }2)= 25\times(\text{number }1)
    i.e. y=25x

    Similarly (\text{number }3)= 2\times(\text{number }1)
    i.e. z=2x

    If you had an equation like: (\text{result 1})=x+3y-2z+xy+z^2 you could use what you know about z and y to change this into
    (\text{result 1})=x+3(25x)-2(2x)+x(25x)+(2x)^2

    Which simplifies to (\text{result 1})=301x+4x^2
    So whatever the formula is you can always simplify it so that it only depends on Number 1

    The formula for result 1 is not simple. Approximately
    (\text{result 1})=0.0005429x^3-0.11158x^2+9.91578x-252.085
    There is some error because I used a computer which gave the answer as a rounded decimal instead of an exact fraction.
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